4.2.1.1. Static Determinacy and Stability of Trusses

Trusses are statically determinate when all the bar forces can be determined from the equations of statics alone. Otherwise, the truss is statically indeterminate.

Static determinacy refers to whether a truss can be analyzed using only the equations of static equilibrium. If every internal force can be calculated solely based on the static conditions (i.e., balance of forces and moments), the truss is termed statically determinate. If there are relationships in the truss that cannot be resolved using these static equations, then it is statically indeterminate. This means we may need more advanced analysis techniques to determine the internal forces.

Imagine a bridge made of trusses. If you can calculate all the forces in the bridge's members simply by measuring the weights and applying basic physics principles, the bridge is statically determinate. But if the bridge is so complex that you need special calculations or simulations to find out the forces, it's statically indeterminate, like a puzzle with pieces that can fit together in various ways.

A truss may be statically/externally determinate or indeterminate with respect to the reactions (more than 3 or 6 reactions in 2D or 3D problems respectively). A truss may be internally determinate or indeterminate.

A truss can be classified in two major ways concerning its reactions and internal members. Externally, a truss is determinate if the number of reactions at the supports is sufficient (3 in 2D, 6 in 3D) to maintain equilibrium. If there are too many reactions, it is over-constrained and unstable. Internally, the truss's members are considered determinate if the total number of members and reactions (m + R) allows for a complete solution with the statics equations available (2j for 2D, 3j for 3D). If the count exceeds what is required, it becomes internally indeterminate.

Think of a game of Jenga. If the base is sturdy enough to support the blocks (like having the right number of reactions), you can take blocks off without causing the tower to collapse (being externally determinate). However, if you add too many blocks or remove others without balancing properly, the tower may fall (externally indeterminate). The same applies to the members: if you have too many members without proper support, it becomes unstable.

If we refer to j as the number of joints, R the number of reactions and m the number of members, then we would have a total of m+R unknowns and 2j (or 3j) equations of statics (2D or 3D at each joint). If we do not have enough equations of statics then the problem is indeterminate; if we have too many equations then the truss is unstable.

In analyzing a truss, you take into account the number of joints (j), reactions (R), and members (m). The unknowns become m + R. Each joint contributes equations based on equilibrium rules, giving 2 equations in 2D or 3 in 3D per joint. If the equations available do not match the unknown forces (too few), then the truss cannot be fully solved (indeterminate). Contrarily, if the equations exceed the unknowns available, it indicates that there is not enough flexibility in the structure, deeming it unstable.

Imagine trying to find out how to balance a scale with weights. Each weight represents the unknown forces in our truss structure. If you only have the scale with two balances (two equations), but you are trying to balance three weights (unknowns), it's impossible to solve because you lack enough tools (equations). On the flip side, if you try to balance excessive weights but your scale can only handle a limited number of positions, it will topple over, similar to an unstable truss.

If m < 2j (in 2D) the truss is not internally stable and it will not remain a rigid body when it is detached from its supports. However, when attached to the supports, the truss will be rigid.

An important condition for internal stability is the relationship between the number of members and joints. If the number of members (m) is less than double the number of joints (2j) in a 2D truss, the structure lacks enough members to maintain rigidity when disconnected from its support. This means that while attached, it may appear stable, but once free, it can collapse or deform significantly.

Consider a spider web. If the web has too few threads connecting the points, it may look okay when hanging, but any small disturbance can rip it apart once one part is detached from its support point. However, when the web is complete with sufficient threads, it maintains its shape and strength in place.

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